The circuit current is 2A. This is found by looking at
R5. Since R5 is connected to the battery in series, it has the full
circuit current. The current splits 3 ways in the top section. The
R1 and R2 section takes 0.5A and the R3 section takes 0.5A. This means
that the R4 branch must take 1A. (0.5A + 0.5A + 1A = 2A, the circuit
current)

Using Ohm's law, R4 = V4/I4 = 10V/1A = 10 ohms.

The voltage supplied by the battery is 50V. The top
section takes 10V and the bottom section takes 20V, this means that R5 has a
voltage drop of 20V. The 3 sections of this circuit are in series with
each other, therefore each must take part of the voltage supplied.

Using Ohm's law, R5 = V5/I5 = 20V/2A = 10ohms.

The current in R7 is the same as the current in R6 since they
are connected in series. Therefore I7 = 1A.